Fastest way to check if two integers are on the same side of 0

Question

I need to check if two integers are on the same side of zero many times. I don\'t care if it\'s positive or negative, just that it\'s the same side... and performance is very im

Answer 1

(int1 ^ int2) >> 31 == 0 ? /*on same side*/ : /*different side*/ ; This doesn't necessarily handle 0 correctly I'm not sure what you wanted to do in that case.
EDIT: also wanted to point out that if this was in c instead of java, it could be optimized further by getting rid of the == 0 because of the way that booleans work in c, the cases would be switched though

Answer 2

I would bitcast them to unsigned int, and xor the MSB (most-significant-bit) - much faster than any comparison (which does a subtraction) or multiplication

Answer 3

if (int1 == 0 || int2 == 0) {
    // handle zero
} else if ((int1 >> 31) == (int2 >> 31)) {
    // same side
} else {
    // different side
}

or

if (int1 == 0 || int2 == 0) {
    // handle zero
} else if ((int1 & Integer.MIN_VALUE) == (int2 & Integer.MIN_VALUE)) {
    // same side
} else {
    // different side
}

The idea of both is the same - strip all but the sign bit, and then compare that for equality. I'm not sure which is faster, the right shift (>>) or the bitwise and (&).

Answer 4

 int int1    = 3;
 int int2    = 4;
 boolean res = ( (int1 * int2) >= 0) ? true : false;

 System.out.println(res);

Answer 5

Another answer...

final int i = int1 ^ int2;

if (i == 0 && int1 == 0) {
    // both are zero
} else if (i & Integer.MIN_VALUE == Integer.MIN_VALUE) {
    // signs differ
} else {
    // same sign
}

Answer 6

Alternate answers

Compare the sign bit

return ((n >> 31) ^ (n2 >> 31) ) == 0 ? /* same */ : /* different */;

Alternate way of comparing sign bit

return (((int1 & 0x80000000) ^ (int2 & 0x80000000))) == 0 ? /* same */ : /* different */;

and I just verified but Op's code is wrong when int1 == int2. The following will always print different if they are the same.

if (int1 == 0 || int2 == 0) {
    // handle zero
} else if ((int1 ^ int2) < 0) {
    // same side
} else {
    // different side
}